Free Body Diagram Problem

Question

I'm having trouble with this problem:

G and a are given, and we are asked to calculate A, B and C.

What I've Done:
The first thing I tried was a free body diagram cutting through A and C to create one body and C and B to create another.

From here I calculated $A_y$ and $C_y$ as being equal to $-G\frac{\sqrt{2}}{4}$
Then I moved onto the second body:

Here I ran into problems immediately. If I placed the moment at either B or C then it follows that the x component of the other must be zero, which it isn't (I have a numerical solution). So I've either made a mistake with my free body diagram or I'm doing something wrong with the second moment.

Here are my equalibrium calculations:
Body 1:
$$\Sigma M^{A}=C_y\cdot 2a+G\frac{\sqrt{2}}{2}=0 \\ \Sigma F_x=A_x-C_x+G\frac{\sqrt{2}}{2}=0 \\ \Sigma F_y=A_y+C_y+G\frac{\sqrt{2}}{2}=0$$

Body 2: $$\Sigma M^{B}=3a\cdot C_x=0 \\ \Sigma F_x=C_x+B_x=0 \\ \Sigma F_y=B_y-C_y=0$$

My results obviously make no sense. What am I doing wrong? I have also tried to calculate for the whole structure but there were too many unknows which I couldn't get rid of.

Answer 1

Let's start by transforming the load $G$ into the resultant forces applied to the structure. At point $R$, we have a vertical component equal to $G$ and an inclined component $G$. Since the inclined component is at 45°, we get a total force equal to $G\left(1 + \dfrac{1}{\sqrt2}\right)$ downwards and $\dfrac{G}{\sqrt2}$ to the left. Relatedly, at the end of the rope (between $A$ and $C$), the concentrated force has upwards vertical and rightward horizontal components both equal to $\dfrac{G}{\sqrt2}$.

The model is therefore equal to (for $G = 100$):

As you already found, we can isolate $AC$. It behaves like a simply supported beam with one horizontal constraint at $A$. Therefore, the horizontal load is fully absorbed by $A_x$. The vertical component is evenly split (due to being at the midspan of $AC$) between $A$ and $C$.

Now we have to deal with the forces at the top of the structure. The vertical force will be fully absorbed by $B$ (but is slightly offset by the vertical component at the end of the rope discussed above which is transmitted to $C$). The horizontal force, however, we can't yet know. For that, we need to balance out the moment around $B$ considering these forces and the horizontal force applied on/by $AC$.

$$\begin{gather}\sum M_B = -G\left(1 + \dfrac{1}{\sqrt2}\right)\cdot3a + \dfrac{G}{\sqrt2}\cdot7a - A_x'\cdot3a = 0 \\ \therefore A_x' = G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} \end{gather}$$

where $A_x'$ is the horizontal reaction in $A$ due to those forces, to be added to the reaction found previously of $\dfrac{G}{\sqrt2}$. So, $B_x = \dfrac{G}{\sqrt2} - A_x'$ (in this case $A_x'$ is negative, so this ends up being a sum).

So we end up with: $$\begin{alignat}{3} A_x &= -\dfrac{G}{\sqrt2} + G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{1}{3\sqrt2} - 1\right) &&\approx -0.764G \\ A_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\ B_x &= \dfrac{G}{\sqrt2} - G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(1 - \dfrac{1}{3\sqrt2}\right) &&\approx 0.764G \\ B_y &= G\left(1 + \dfrac{1}{\sqrt2}\right) - \dfrac{G}{2\sqrt2} &&= G\left(1 + \dfrac{1}{2\sqrt2}\right) &&\approx 1.354G \\ C_x &= G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{4}{3\sqrt2} - 1\right) = &&\approx -0.057G \\ C_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\ \end{alignat}$$

And now, to check our work:

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Alternatively, using your method, what you forgot to do was to consider the bending moment due to the forces at the top of the structure in your second cut ($BC$). That moment is equal to $G\left(1 + \dfrac{1}{\sqrt2}\right)3a - \dfrac{G}{\sqrt2}\cdot4a = Ga\left(3 - \dfrac{1}{\sqrt2}\right)$, which can only be balanced by the force binary generated by $A_x$ and $B_x$. Dividing the moment by the $3a$ lever arm between $A_x$ and $B_x$, we get that each of those forces must be equal to $\pm\dfrac{Ga\left(3 - \dfrac{1}{\sqrt2}\right)}{3a} = \pm G\left(1 - \dfrac{1}{3\sqrt2}\right)$, which is precisely what we got for $A_x$ and $B_x$ above. Knowing this, you can then solve for the other variables as well.

_{All diagrams given by Ftool, a free 2D frame analysis tool.}

Answer 2

The balance equations would be the following

$$\Sigma F_x = A_x-C_x+\frac{G}{\sqrt{2}}=0$$ $$\Sigma F_y = A_y+C_y+\frac{G}{\sqrt{2}}=0$$ $$\Sigma M_P = A_y a-C_y a=0$$ $$\Sigma F_x = B_x+C_x-\frac{G}{\sqrt{2}}=0$$ $$\Sigma F_y = B_y-C_y-\left(\frac{G}{\sqrt{2}}+G\right)=0$$ $$\Sigma M_B = C_x (3 a)-\frac{G}{\sqrt{2}}(7 a)+ \left(\frac{G}{\sqrt{2}}+G\right)(3 a)=0$$

And the solution turns out as $$ A_x=\frac{1}{6} \left(\sqrt{2} G-6 G\right) \ \ A_y=-\frac{G}{2 \sqrt{2}}$$ $$B_x=\frac{1}{6} \left(6 G-\sqrt{2} G\right)\ \ B_y=\frac{1}{4} \left(\sqrt{2} G+4 G\right)$$ $$C_x=\frac{1}{3} \left(2 \sqrt{2} G-3 G\right) \ \ C_y=-\frac{G}{2 \sqrt{2}}$$

The free-body diagrams